A study was done on proctored and nonproctored tests. The results are shown in t

Question

A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. a. Use a 0.01 significance level, and test the claim that students taking nonproctored tests gel a higher mean score than those taking proctored tests. What are the null and alternative hypotheses? H_0: mu_1 = mu_2 H_1: mu_1 mu_2 H_0: mu_1 = mu_2 H_: mu_1 notequalto mu_2H_0: mu_1 notequalto mu_2 H_1: mu_1

Explanation / Answer

Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate, non proctored get higher mean than proctored - H1: u1 < u2
Test Statistic
X(Mean)=75.64
Standard Deviation(s.d1)=10.76 ; Number(n1)=35
Y(Mean)=84.15
Standard Deviation(s.d2)=20.97; Number(n2)=31
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =75.64-84.15/Sqrt((115.7776/35)+(439.7409/31))
to =-2.035
| to | =2.035
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 30 d.f is 2.457
We got |to| = 2.03468 & | t | = 2.457
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Left Tail - Ha : ( P < -2.0347 ) = 0.0254
Hence Value of P0.01 < 0.0254,Here We Do not Reject Ho

[ANSWERS]
a. Ho: u1 = u2 , H1: u1 < u2
b. to =-2.035, P - value 0.0254
c. Fail to reject H0. There is not sufficient evidence to su’port the cI3irn that students taing
nonproctored tests get a higher mean score than those taking proctored

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